2013年山东省第四届ACM大学生程序设计竞赛E题:Alice and Bob

题目描述

Alice and Bob like playing games very much.Today, they introduce a new game.

There is a polynomial like this: (a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1). Then Alice ask Bob Q questions. In the expansion of the Polynomial, Given an integer P, please tell the coefficient of the x^P.

Can you help Bob answer these questions?

输入

The first line of the input is a number T, which means the number of the test cases.

For each case, the first line contains a number n, then n numbers a0, a1, .... an-1 followed in the next line. In the third line is a number Q, and then following Q numbers P.

1 <= T <= 20

1 <= n <= 50

0 <= ai <= 100

Q <= 1000

0 <= P <= 1234567898765432

输出

For each question of each test case, please output the answer module 2012.

示例输入

1
2
2 1
2
3
4

示例输出

2
0

提示

The expansion of the (2*x^(2^0) + 1) * (1*x^(2^1) + 1) is 1 + 2*x^1 + 1*x^2 + 2*x^3

 

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分析：这是一道数学题。简单说来就是给出一堆式子，求x^P 的系数。求公比为2的首项是1的等比数列的和，得到P 的最大值为max(P) = (1 - 2^50) / (1 - 2) = 1125899906842623，而题目中给出P 的输入最大值inputMax(P)为1234567898765432。因此inputMax(P) < 2^51(这保证了下面的程序不会出现数组下标越界的问题)。另外，观察式子不难得出x^P的系数为P 的二进制形式中为1 的部分对应的ai 的乘积。例如对X^11，11写成2进制为(1011)₂，那么其系数就是a0*a1*a3。程序如下：

#include<stdio.h>
#include<string.h>
#define MAXN 50
#define MOD 2012
int a[MAXN+];
int main(void)
{
    int t, n, q, index, ans;
    long long int exp; // 注意定义成int会WA, 因为据题意exp可达1234567898765432这么大

    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        memset(a, , sizeof(a));
        for(int i = ; i <= n; i++)
            scanf("%d", &a[i]);

        scanf("%d", &q);
        while(q--)
        {
            scanf("%lld", &exp); // SDUT: C/C++评测程序只支持long long, 不支持int64

            ans = index = ;
            while(exp)
            {
                if(exp %  != )
                    ans = (ans*a[index]) % MOD;
                index++;
                exp /= ;
            }

            printf("%d\n", ans);
        }

    }

    return ;
}